Answer
$f(x)=(x+1)(x+3)(x-2)$
Zeros: $-3,\ \ -1,\ \ 2,$ and all with multiplicity $1$.
Work Step by Step
Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$.
We see from the given polynomial function that it has at most $3$ real zeros as the degree is $3$.
The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 3, \pm 6$ and $n=\pm 1$
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm 2, \pm 3, \pm 6$
We test with synthetic division. We will try $x+1$.
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &2 &-5&-6\\\hline
&-1&-1 &6\\\hline
1&1 &-6& |\ \ 0\end{array}$
Thus, we have: $f(x)=(x+1)(x^{2}+x-6) \\=(x+1)(x+3)(x-2)$
Zeros: $-3,\ \ -1,\ \ 2,$ and all with multiplicity $1$.
